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A nucleus with Z=92 emits the following in a sequence $α, β^{-}, β^{-}, α, α, α, β^{-}, β^{-}, α, β^{+}, β^{+}$ and $α$ . The atomic number of the final element will be

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Correct option is D

Here 6α,4β−,2β+ are particles emitted, thus new atomic number is Z=92−[62-4+2] ⇒Z=82

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A nucleus with Z=92 emits the following in a sequenceα,β−,β−,α,α,α,β−,β−,α,β+,β+ and α . The atomic number of the final element will be

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A nucleus with Z=92 emits the following in a sequence $α, β^{-}, β^{-}, α, α, α, β^{-}, β^{-}, α, β^{+}, β^{+}$ and $α$ . The atomic number of the final element will be